Difference between revisions of "Talk:1516: Win by Induction"
(Doubling does not make it uncountably infinite.) |
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Is the alt text a reference to double-yolkers (eggs with two yolks)? [http://www.bbc.co.uk/news/magazine-16118149 They're only about 1 in every 1000] but it seems like an obvious reference. --[[User:Fenn|Fenn]] ([[User talk:Fenn|talk]]) 08:32, 24 April 2015 (UTC) | Is the alt text a reference to double-yolkers (eggs with two yolks)? [http://www.bbc.co.uk/news/magazine-16118149 They're only about 1 in every 1000] but it seems like an obvious reference. --[[User:Fenn|Fenn]] ([[User talk:Fenn|talk]]) 08:32, 24 April 2015 (UTC) | ||
:Makes sense to me. I didn't even think of double yolks until you mentioned it here. [[Special:Contributions/173.245.50.89|173.245.50.89]] 09:04, 24 April 2015 (UTC)BK201 | :Makes sense to me. I didn't even think of double yolks until you mentioned it here. [[Special:Contributions/173.245.50.89|173.245.50.89]] 09:04, 24 April 2015 (UTC)BK201 | ||
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| + | The explanation currently says that doubling makes it uncountably infinite. I'm pretty sure that doubling at each step (or every few steps) is still a countable infinite set. Proof here: http://practicaltypography.com/the-infinite-pixel-screen.html (see section "The internet demands a recount", because the first attempt is wrong). We can also prove it using the same argument as when proving that N x N is countable infinite (making zig-zag), but in this case making a breadth-first search of the tree of Pikachus: map 1 to the first Pikachu, map 2 and 3 to the two Pikachus at the second level, map 4, 5, 6, 7 to the four Pikachus at the third level, map (2^(n-1))…((2^n) - 1) to the 2^(n-1) Pikachus at level n. | ||
Revision as of 09:21, 24 April 2015
Is the alt text a reference to double-yolkers (eggs with two yolks)? They're only about 1 in every 1000 but it seems like an obvious reference. --Fenn (talk) 08:32, 24 April 2015 (UTC)
- Makes sense to me. I didn't even think of double yolks until you mentioned it here. 173.245.50.89 09:04, 24 April 2015 (UTC)BK201
The explanation currently says that doubling makes it uncountably infinite. I'm pretty sure that doubling at each step (or every few steps) is still a countable infinite set. Proof here: http://practicaltypography.com/the-infinite-pixel-screen.html (see section "The internet demands a recount", because the first attempt is wrong). We can also prove it using the same argument as when proving that N x N is countable infinite (making zig-zag), but in this case making a breadth-first search of the tree of Pikachus: map 1 to the first Pikachu, map 2 and 3 to the two Pikachus at the second level, map 4, 5, 6, 7 to the four Pikachus at the third level, map (2^(n-1))…((2^n) - 1) to the 2^(n-1) Pikachus at level n.
